The scalar field $f(x, y) = x\ln(y^2) - x$ has a critical point at $(0, \sqrt{e})$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D, Checked) D The test is inconclusive
Answer: The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= \ln(y^2) - 1 \\ \\ f_y &= \dfrac{x}{y^2}(2y) = \dfrac{2x}{y} \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= 0 \\ \\ f_{yx} &= \dfrac{2y}{y^2} = \dfrac{2}{\sqrt{e}} \\ \\ f_{xy} &= \dfrac{2}{y} = \dfrac{2}{\sqrt{e}} \\ \\ f_{yy} &= \dfrac{-2x}{y^2} = 0 \end{aligned}$ Therefore: $H = (0)(0) - \left( \dfrac{2}{\sqrt{e}} \right) \left( \dfrac{2}{\sqrt{e}} \right) = -\dfrac{4}{e}$ Because $H$ is negative, we know that the critical point is a saddle point.